# What is the slope of the tangent line of sqrt(y-e^(x-y))= C , where C is an arbitrary constant, at (-2,1)?

Mar 9, 2018

Equation of tangent is $y - 1 = \frac{1}{{e}^{3} + 1} \left(x + 2\right)$

#### Explanation:

As tangent is saught at the point $\left(- 2 , 1\right)$, it is apparent that the point lies on the curve $\sqrt{y - {e}^{x - y}} = C$, we have

$\sqrt{1 - {e}^{- 2 - 1}} = C$

i.e. $C = \sqrt{1 - \frac{1}{e} ^ 3}$

Hence function is $\sqrt{y - {e}^{x - y}} = \sqrt{1 - \frac{1}{e} ^ 3}$

or $y - {e}^{x - y} = 1 - \frac{1}{e} ^ 3$

As slope of tangent is the value of first derivative at the point, here $\left(- 2 , 1\right)$, let us find first derivative by implicit differentiation. Differentiating $y - {e}^{x - y} = 1 - \frac{1}{e} ^ 3$, we have

$\frac{\mathrm{dy}}{\mathrm{dx}} - {e}^{x - y} \left(1 - \frac{\mathrm{dy}}{\mathrm{dx}}\right) = 0$

or $\frac{\mathrm{dy}}{\mathrm{dx}} \left(1 + {e}^{x - y}\right) = {e}^{x - y}$

i.e. $\frac{\mathrm{dy}}{\mathrm{dx}} = {e}^{x - y} / \left(1 + {e}^{x - y}\right)$

Hence, slope of tangent is ${e}^{- 3} / \left(1 + {e}^{- 3}\right)$ or $\frac{1}{{e}^{3} + 1}$

and equation of tangent is $y - 1 = \frac{1}{{e}^{3} + 1} \left(x + 2\right)$

graph{(sqrt(y-e^(x-y))-sqrt(1-1/e^3))(x-(e^3+1)y+3+e^3)=0 [-11.29, 8.71, -4.52, 5.48]}